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If y = cos(kt), then its first two derivatives are
y' = -k sin(kt)
y'' = -k² cos(kt)
Substituting y and y'' into 49y'' = -16y gives
-49k² cos(kt) = -15 cos(kt)
⇒ 49k² = 15
⇒ k² = 15/49
⇒ k = ±√15/7
Note that both values of k give the same solution y = cos(√15/7 t) since cosine is even.
Answer:
15
-----
32
Step-by-step explanation:
3 5 15
-- * -- == ---
4 8 32
Answer:
coefficient of the term of degree 4 in this polynomial is 3
Step-by-step explanation:
